"amara vati" <amaraavati@yahoo.com> wrote in message
news:f89b870.0408110653.35703f9c@posting.google.com...
> a zero mean process could have no DC component in it. PSD is a
> statitistical curve and not the spectrum itself. PSD non zero at DC
> doesnt meant there is a net DC component in noise. If that be the
> case, I should be able to generate power from noise. To understand PSD
> better, you should see that if u integrate PSD over the entire
> spectrum, u get the variance of the process. that is PSD gives you a
> picture of how much each part of the spectrum contrubutes to the
> variance. that means you should add noise of same variance to all the
> tones as white noise has uniform PSD
>
> amar
>
> le.com>...
> > White noise has a PSD = No/2 for all frequencies w. Does this hold
> > true at w = 0? In this case does that mean that white noise must have
> > a DC component. What about zero mean white noise? Does this have:
> > PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta?
> >
> > I am simulating a baseband OFDM system and I want to add system noise
> > at the appropriate level for a given SNR. Do I add noise on all tones
> > equally? Given that the DC tone is not used, should I also turn off
> > the noise on this tone? I am assuming a zero-mean white-noise channel.

If you pass theoretical white noise through a simple RC time-constant you
can calculate the theoretical
PSD at the output quite easily and it goes right down to DC of course.
However this does not mean there is a DC component at the output.As for the
white noise itself,if it has zero mean then there is no DC component.With AC
coupling this is usually the case unless there are DC offsets somewhere.


Tom

Tim Wescott wrote:
> Andor wrote:
...
> It means that in theory there will be no shot where the exact centroid 
> of the bullet lies over the exact centroid of the y axis -- the same for 
> any other finite set of points on the target that have zero area.

You argue as follows: The probability of the shot landing on the Y axis 
is zero, therefore "in theory", this will never happen.

I argue: The probability of the shot landing anywhere is zero (given 
2-dim Gaussian distribution). Still, every shot hits some coordinate 
pair (X,Y). So your reasoning that an event with zero probability cannot 
happen, is flawed.


> What this tells us about the DC content of the Gaussian white noise is 
> that because DC is a single point on the frequency axis when you 
> integrate the PDF from 0 to 0 there will be no area under the curve 
> (finite number * zero length = zero).

Agreed, but that is not what I was aiming at.

> If you mean n as being sample time then s(n) = 1 for all n has a mean 
> value of 1, not zero, so it cannot be an instance of a zero mean 
> Gaussian white noise process.  

But of course it can. It just has zero probability - but as we saw 
above, that does not immediately imply that it cannot happen.

The law of large numbers tells us that the mean of a standard Gaussian 
white noise process with time index n converges to zero P-almost-surely 
as n goes to infinity, where P is the probability measure implied by the 
Gaussian density. This says that the set where the mean does not 
converge to zero has no mass with regard to measure P. Above example 
(with the marksman) shows that a set with zero mass need not necessarily 
be empty (for exactly that P). The process s(n) = 1, for all n, is in 
such a set.


Regards,
Andor

amaraavati@yahoo.com (amara vati) writes:

> after some thought, I feel I could be wrong. I am confused. cud
> somebody enliten pls.
>
> amar

Hi Amar,

You've already gotten some good responses, but allow me to reason with you 
from this viewpoint.

Define energy over a period T as 

  E(T) = /int_{-T/2}^{+T/2} |x(t)|^2 dt.

Note that, unit-wise, this integral works out since |x(t)|^2 has units of
power (assuming x is volts and there is an implied resistance of 1 ohm),
and power times time (|x(t)|^2 * dt) is energy. 

A finite-energy signal is one in which 

  \lim_{T --> \infty} E(T) < \infty.

Then, if a signal has finite-energy, it must have zero power, since the total 
average power in a signal is \lim_{T --> \infty} E(T)/T. Since E(T) is always
finite but T increases without bound, the ratio must approach 0.

Similarly, define the average power over a period T as 

  P(T) = (1/T) * E(T).

A finite-power signal is one in which 

  \lim_{T --> \infty} P(T) < \infty.

Note that a finite non-zero power signal must necessarily have infinite
energy, otherwise the limit above would tend toward zero since T would
keep increasing while E(T) stops at some finite value.

Now also note that power spectral density has the units of watts/Hz,
which is equivalent to joules since watts = joules/second and Hz = 1/second.

OK, so what? Well, this means that any signal x which has a PSD Sxx(w)
(I'm using "w" here for omega, 2*pi*f) that is non-zero somewhere 
has a finite, non-zero power and thus must have infinite energy.

That infinite energy could come from a single sinusoid at frequency
w0, in which case Sxx(w0) = A * \delta(w-w0), where "\delta(x)" is the
Dirac delta function. This must be true since there is only one
isolated point in the Sxx(w) function that has a non-zero value, thus
the energy represented by that point must be infinite. Stated another
way, the power in a zero-Hz bandwidth must be non-zero.

Or the infinite energy could come from a continuous range of
frequencies, say, w1 to w2. In that case, the energy at one particular
frequency can be finite, while the sum of energies at an infinite
number of points (those in the interval of w of w1 to w2) remains
infinite. The power, then, at any one frequency in the range w1 to
w2 is zero since the energy is finite. 

So, that explains it. A signal with a smooth PSD (i.e., no Dirac
delta functions) will have a finite energy at any particular
frequency, which means the power at that frequency is zero. For
a white spectrum, there is energy at DC, but there is no power
there.
-- 
%  Randy Yates                  % "Bird, on the wing,
%% Fuquay-Varina, NC            %   goes floating by
%%% 919-577-9882                %   but there's a teardrop in his eye..."
%%%% <yates@ieee.org>           % 'One Summer Dream', *Face The Music*, ELO
http://home.earthlink.net/~yatescr

Andor wrote:

> Jerry Avins wrote:
> ...
> 
>>It will help if you clarify your understanding of what PDF, PSD, and
>>related quantitise mean. Ignore the acronyms; consider a marksman aiming
>>at a piece of graph paper. His aim is perturbed by so many independent
>>disturbances that the deviation of his shots is Gaussian. He is a very
>>good marksman, so the centroid of his cluster is right at his point of
>>aim: the origin. Consider now the probability of a shot falling at some
>>X location (ignore Y position) and draw this curve. Smooth it to
>>indicate what it would look like if he had fired a very large number of
>>shots. I think you can see that it would be symmetric about the Y axis,
>>have its greatest value at X=0, and continue out very far at low
>>amplitude. In short, a classical bell curve. Reading the curve, what is
>>the probability that a shot will fall precisely on the Y axis? Zero, of
>>course. How can that be?
> 
> 
> Does that mean no shot will ever fall on the Y axis? 

I don't address that. My intention is to consider (plot) only the X 
value of position, ignoring Y.


> Or conversly: A
> shot entered the X axis at, say, point 0.5 - what is the probability
> of that? And what does this tell us about DC content of Gaussian white
> noise?

I think the OP's error arose from failing to distinguish probability
from probability density. I tried to stimulate an example that requires
the distinction.

> Consider the signal s(n) = 1 for all n. Could this be an instance of
> zero mean Gaussian white noise process? If so, how does it differ (in
> probability) from other particular instance where the mean converges
> to zero as n -> infinity?

Jerry
-- 
... the worst possible design that just meets the specification - almost
a definition of practical engineering.                     .. Chris Bore
������������������������������������������������������������������������

Andor wrote:

> Jerry Avins wrote:
> ...
> 
>>It will help if you clarify your understanding of what PDF, PSD, and
>>related quantitise mean. Ignore the acronyms; consider a marksman aiming
>>at a piece of graph paper. His aim is perturbed by so many independent
>>disturbances that the deviation of his shots is Gaussian. He is a very
>>good marksman, so the centroid of his cluster is right at his point of
>>aim: the origin. Consider now the probability of a shot falling at some
>>X location (ignore Y position) and draw this curve. Smooth it to
>>indicate what it would look like if he had fired a very large number of
>>shots. I think you can see that it would be symmetric about the Y axis,
>>have its greatest value at X=0, and continue out very far at low
>>amplitude. In short, a classical bell curve. Reading the curve, what is
>>the probability that a shot will fall precisely on the Y axis? Zero, of
>>course. How can that be?
> 
> 
> Does that mean no shot will ever fall on the Y axis? Or conversly: A
> shot entered the X axis at, say, point 0.5 - what is the probability
> of that? And what does this tell us about DC content of Gaussian white
> noise?
> 
> Consider the signal s(n) = 1 for all n. Could this be an instance of
> zero mean Gaussian white noise process? If so, how does it differ (in
> probability) from other particular instance where the mean converges
> to zero as n -> infinity?
> 
> 
>>Jerry

It means that in theory there will be no shot where the exact centroid 
of the bullet lies over the exact centroid of the y axis -- the same for 
any other finite set of points on the target that have zero area.

What this tells us about the DC content of the Gaussian white noise is 
that because DC is a single point on the frequency axis when you 
integrate the PDF from 0 to 0 there will be no area under the curve 
(finite number * zero length = zero).

If you mean n as being sample time then s(n) = 1 for all n has a mean 
value of 1, not zero, so it cannot be an instance of a zero mean 
Gaussian white noise process.  If you were on the grounds of a 
University math department you could claim that it's a Gaussian process 
with mean = 1 and sigma = 0, but that's degenerate.

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Jerry Avins wrote:
...
> It will help if you clarify your understanding of what PDF, PSD, and
> related quantitise mean. Ignore the acronyms; consider a marksman aiming
> at a piece of graph paper. His aim is perturbed by so many independent
> disturbances that the deviation of his shots is Gaussian. He is a very
> good marksman, so the centroid of his cluster is right at his point of
> aim: the origin. Consider now the probability of a shot falling at some
> X location (ignore Y position) and draw this curve. Smooth it to
> indicate what it would look like if he had fired a very large number of
> shots. I think you can see that it would be symmetric about the Y axis,
> have its greatest value at X=0, and continue out very far at low
> amplitude. In short, a classical bell curve. Reading the curve, what is
> the probability that a shot will fall precisely on the Y axis? Zero, of
> course. How can that be?

Does that mean no shot will ever fall on the Y axis? Or conversly: A
shot entered the X axis at, say, point 0.5 - what is the probability
of that? And what does this tell us about DC content of Gaussian white
noise?

Consider the signal s(n) = 1 for all n. Could this be an instance of
zero mean Gaussian white noise process? If so, how does it differ (in
probability) from other particular instance where the mean converges
to zero as n -> infinity?

> 
> Jerry

amara vati wrote:
> after some thought, I feel I could be wrong. I am confused. cud
> somebody enliten pls.

Enlighten or more likely, further confuse.

 From a probability perspective, think of constant voltage drawn from 
some sort of Gaussian constant voltage generator where the mean is zero 
and the variance is 1 volt.  The actual voltage you measure is not 
likely to be zero, but if you repeat the experiment enough times the 
average of all the measurements will tend toward zero as you continue to 
  repeat the experiment.

> 
> amar
> 
> porterboy76@yahoo.com (porterboy) wrote in message news:<c4b57fd0.0408110121.55376b70@posting.google.com>...
> 
>>White noise has a PSD = No/2 for all frequencies w. Does this hold
>>true at w = 0? In this case does that mean that white noise must have
>>a DC component. What about zero mean white noise? Does this have:
>>PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta?
>>
>>I am simulating a baseband OFDM system and I want to add system noise
>>at the appropriate level for a given SNR. Do I add noise on all tones
>>equally? Given that the DC tone is not used, should I also turn off
>>the noise on this tone? I am assuming a zero-mean white-noise channel.

porterboy wrote:

> White noise has a PSD = No/2 for all frequencies w. Does this hold
> true at w = 0? In this case does that mean that white noise must have
> a DC component. What about zero mean white noise? Does this have:
> PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta?
> 
> I am simulating a baseband OFDM system and I want to add system noise
> at the appropriate level for a given SNR. Do I add noise on all tones
> equally? Given that the DC tone is not used, should I also turn off
> the noise on this tone? I am assuming a zero-mean white-noise channel.

It will help if you clarify your understanding of what PDF, PSD, and
related quantitise mean. Ignore the acronyms; consider a marksman aiming
at a piece of graph paper. His aim is perturbed by so many independent
disturbances that the deviation of his shots is Gaussian. He is a very
good marksman, so the centroid of his cluster is right at his point of
aim: the origin. Consider now the probability of a shot falling at some
X location (ignore Y position) and draw this curve. Smooth it to
indicate what it would look like if he had fired a very large number of
shots. I think you can see that it would be symmetric about the Y axis,
have its greatest value at X=0, and continue out very far at low
amplitude. In short, a classical bell curve. Reading the curve, what is
the probability that a shot will fall precisely on the Y axis? Zero, of
course. How can that be?

Jerry
-- 
... the worst possible design that just meets the specification - almost
a definition of practical engineering.                     .. Chris Bore
������������������������������������������������������������������������

after some thought, I feel I could be wrong. I am confused. cud
somebody enliten pls.

amar

porterboy76@yahoo.com (porterboy) wrote in message news:<c4b57fd0.0408110121.55376b70@posting.google.com>...
> White noise has a PSD = No/2 for all frequencies w. Does this hold
> true at w = 0? In this case does that mean that white noise must have
> a DC component. What about zero mean white noise? Does this have:
> PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta?
> 
> I am simulating a baseband OFDM system and I want to add system noise
> at the appropriate level for a given SNR. Do I add noise on all tones
> equally? Given that the DC tone is not used, should I also turn off
> the noise on this tone? I am assuming a zero-mean white-noise channel.

a zero mean process could have no DC component in it. PSD is a
statitistical curve and not the spectrum itself. PSD non zero at DC
doesnt meant there is a net DC component in noise. If that be the
case, I should be able to generate power from noise. To understand PSD
better, you should see that if u integrate PSD over the entire
spectrum, u get the variance of the process. that is PSD gives you a
picture of how much each part of the spectrum contrubutes to the
variance. that means you should add noise of same variance to all the
tones as white noise has uniform PSD

amar

le.com>...
> White noise has a PSD = No/2 for all frequencies w. Does this hold
> true at w = 0? In this case does that mean that white noise must have
> a DC component. What about zero mean white noise? Does this have:
> PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta?
> 
> I am simulating a baseband OFDM system and I want to add system noise
> at the appropriate level for a given SNR. Do I add noise on all tones
> equally? Given that the DC tone is not used, should I also turn off
> the noise on this tone? I am assuming a zero-mean white-noise channel.